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Two couples and one single person are seated at random in a

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Two couples and one single person are seated at random in a row of five chairs. What is the probability that neither of the couples sits together in adjacent chairs?

A. 1/5
B. 1/4
C. 3/8
D. 2/5
E. 1/2

Open discussion of this question is here: two-couples-and-one-single-person-are-seated-at-random-in-a-125807.html. In case of any question please post in that thread.
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New post 30 Nov 2007, 11:27
my answer was 4/5. Since it's not in the answer choices, I did a search and fond this post at scorechase:
http://www.scorechase.com/gmat/showthread.php?p=40255

Check out shanky baba's answer towards the end.

Can anyone explain when to use xCx and when not to? I'm a little bit confused with that. Thanks!
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New post 30 Nov 2007, 12:01
Expert's post
d.

let x - a single person, 0 - a person of couples. 1 - a person of first couples, 1 - a person of second couples

x0000 16 cases to seat together at least one couple. (x1122 - 4,x2112 - 4,x2211 - 4, 1221 - 4)
0x000 16 cases to seat together at least one couple.
00x00 8 cases to seat together at least one couple.
000x0 and 0000x 16*2 due to symmetry.

Therefore,

p=1-(16*4+8)/5!=1-8*9/5*4*3*2=1-3/5=2/5

But I've wasted a lot of time to find the way. :cry:

are there any shortcuts?
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New post 29 Dec 2007, 19:03
It looks like not many people will be able to solve this question in under 3 minutes...I'm happy to guess wrong on this on the real test!
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Re: PS: Probability [#permalink]

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New post 26 Aug 2008, 22:41
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tarek99 wrote:
Two couples and one single person are seated at random in a row of five chairs. What is the probability that neither of the couples sits together in adjacent chairs?

a) 1/5
b) 1/4
c) 3/8
d) 2/5
e) 1/2


Please show the steps


P for both couples sit together (C1C1)(C2C2)(S)
= 2!2!3! =24
P for first couple sit together and other couples not (C1C1)(C2C2S)
= 2!4! - 2!2!3! =24
similarly for P second couple sit together and others couples not
=24

\(p=1- (24+24+24)/120 = 1-3/5=2/5\)
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walker wrote:
d.

let x - a single person, 0 - a person of couples. 1 - a person of first couples, 1 - a person of second couples

x0000 16 cases to seat together at least one couple. (x1122 - 4,x2112 - 4,x2211 - 4, 1221 - 4)
0x000 16 cases to seat together at least one couple.
00x00 8 cases to seat together at least one couple.
000x0 and 0000x 16*2 due to symmetry.

Therefore,

p=1-(16*4+8)/5!=1-8*9/5*4*3*2=1-3/5=2/5

But I've wasted a lot of time to find the way. :cry:

are there any shortcuts?


Here my take.. seems short to me.

Consider first couple as a unit and so sits together. The # of arrangements 2*4! (2 ways for couple, 4! for the couple as a unit and the remaining 3 ppl)

Consider second couple as a unit and so sits together. The # of arrangements 2*4! (2 ways for couple, 4! for the couple as a unit and the remaining 3 ppl).

Consider each couple as a unit. The # of arrangements 2*2*3! (2 ways for each couple and 3! for couples a units and the loner)

So total number of ways in which atleast one couple sits together is 2*4! + 2*4! - 2*2*3! (AUB = A+B-AB)

Total number of possible ways of seating 5!

Prob that no couple sits togehter = 1 - (4*4! - 4*3!)/5! = 2/5
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New post 27 Aug 2008, 06:08
bhushangiri wrote:
walker wrote:
d.

let x - a single person, 0 - a person of couples. 1 - a person of first couples, 1 - a person of second couples

x0000 16 cases to seat together at least one couple. (x1122 - 4,x2112 - 4,x2211 - 4, 1221 - 4)
0x000 16 cases to seat together at least one couple.
00x00 8 cases to seat together at least one couple.
000x0 and 0000x 16*2 due to symmetry.

Therefore,

p=1-(16*4+8)/5!=1-8*9/5*4*3*2=1-3/5=2/5

But I've wasted a lot of time to find the way. :cry:

are there any shortcuts?


Here my take.. seems short to me.

Consider first couple as a unit and so sits together. The # of arrangements 2*4! (2 ways for couple, 4! for the couple as a unit and the remaining 3 ppl)

Consider second couple as a unit and so sits together. The # of arrangements 2*4! (2 ways for couple, 4! for the couple as a unit and the remaining 3 ppl).

Consider each couple as a unit. The # of arrangements 2*2*3! (2 ways for each couple and 3! for couples a units and the loner)

So total number of ways in which atleast one couple sits together is 2*4! + 2*4! - 2*2*3! (AUB = A+B-AB)

Total number of possible ways of seating 5!

Prob that no couple sits togehter = 1 - (4*4! - 4*3!)/5! = 2/5


I took <1m too.. :lol:
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New post 27 Aug 2008, 06:34
bhushangiri wrote:
walker wrote:
d.

let x - a single person, 0 - a person of couples. 1 - a person of first couples, 1 - a person of second couples

x0000 16 cases to seat together at least one couple. (x1122 - 4,x2112 - 4,x2211 - 4, 1221 - 4)
0x000 16 cases to seat together at least one couple.
00x00 8 cases to seat together at least one couple.
000x0 and 0000x 16*2 due to symmetry.

Therefore,

p=1-(16*4+8)/5!=1-8*9/5*4*3*2=1-3/5=2/5

But I've wasted a lot of time to find the way. :cry:

are there any shortcuts?


Here my take.. seems short to me.

Consider first couple as a unit and so sits together. The # of arrangements 2*4! (2 ways for couple, 4! for the couple as a unit and the remaining 3 ppl)

Consider second couple as a unit and so sits together. The # of arrangements 2*4! (2 ways for couple, 4! for the couple as a unit and the remaining 3 ppl).

Consider each couple as a unit. The # of arrangements 2*2*3! (2 ways for each couple and 3! for couples a units and the loner)

So total number of ways in which atleast one couple sits together is 2*4! + 2*4! - 2*2*3! (AUB = A+B-AB)

Total number of possible ways of seating 5!

Prob that no couple sits togehter = 1 - (4*4! - 4*3!)/5! = 2/5


Are you able to solve it within 2-3 minutes?
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Re: PS: Probability [#permalink]

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New post 27 Sep 2009, 22:35
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Two couples and one single person are seated at random in a row of five chairs. What is the probability that neither of the couples sits together in adjacent chairs?

a) 1/5
b) 1/4
c) 3/8
d) 2/5
e) 1/2


Soln:
Probability that none will sit together is
= (8 + 8 + 16 + 8 + 8)/120
= 48/120
= 2/5

Ans is D
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Re: PS: Probability [#permalink]

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New post 02 May 2011, 20:20
interesting way to solve this.
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Re: PS: Probability   [#permalink] 02 May 2011, 20:20
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